Answer
$x\tan x-\ln|\sec x|-\frac{x^{2}}{2}+C$
Work Step by Step
$\int x \tan^{2}x dx=x\int\tan^{2}x dx-\int[\frac{d(x)}{dx}\int\tan^{2}x dx]dx$
$=x\int(\sec^{2}x-1)dx-\int[\int(\sec^{2}x-1)dx]dx$
$=x(\tan x-x)-\int(\tan x-x)dx$
$=x\tan x-x^{2}-\ln|\sec x|+\frac{x^{2}}{2}+C$
$=x\tan x-\ln|\sec x|-\frac{x^{2}}{2}+C$
